Question

this is the graph of a function.
choose the graph of its derivative from among the following functions.
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Explanation:

Step1: Analyze the original function's graph

The original function has a local maximum at \( x = 2 \) (since the graph reaches a peak there) and a local minimum? Wait, no, looking at the original graph: on the left, it comes from the top left, goes down, then on the right side of \( x = 0 \), it rises to a peak at \( x = 2 \), then falls. Also, the left part (for \( x < - 4 \) maybe?) and the right part (for \( x > 3 \)) go down. Wait, actually, the original function's derivative will tell us about the slope.

- When the original function is decreasing (slope negative), the derivative is negative.
- When the original function is increasing (slope positive), the derivative is positive.
- At the peak (local maximum) of the original function (at \( x = 2 \)), the derivative should be zero (horizontal tangent).
- Let's track the slope:
- For \( x < - 4 \): The original function is decreasing (going down as \( x \) increases towards -4), so derivative is negative.
- Between \( - 4 \) and \( 2 \): Wait, no, looking at the original graph: from \( x = - 5 \) to \( x = 2 \), wait, no, the graph on the left (left of \( x = 0 \)): at \( x = - 4 \), it crosses the x-axis, then goes down to a minimum? Wait, maybe better to look at the critical points. The original function has a local maximum at \( x = 2 \) (where the slope is zero, derivative zero). Also, let's see the behavior:

- As \( x \) approaches \( -\infty \), the original function is decreasing (slope negative), then at some point (left of \( x = - 4 \)) it starts decreasing? Wait, maybe the original function is a cubic-like? No, the original graph: left side (x < -4) is decreasing (slope negative), then between -4 and 2, wait, no, the graph on the right of x=0: from x=0 to x=2, it's increasing (slope positive), then from x=2 to x=4, it's decreasing (slope negative). Also, on the left of x=0: from x=-5 to x=0, the function is decreasing (slope negative) until some point, then? Wait, maybe I messed up. Let's look at the derivative graphs.

Wait, the bottom - most graph (the last one, the fourth option) is a parabola opening downward, with roots at x = -2 and x = 2? Wait, no, the fourth graph (the one with peak at x=0, crossing x-axis at x=-2 and x=2). Wait, no, let's check the options:

First option: a parabola opening upward, with vertex at (0, -4), crossing x-axis at x=-2 and x=2.

Second option: same as original graph? No, second option looks like the original graph but maybe shifted? No, second option has a peak at x=2, similar to original.

Third option: a parabola opening upward, with vertex at (2, -2), crossing x-axis at x=1 and x=3? Wait, no, third option: on the left, it's increasing (slope positive) for x < -4, then decreasing, then increasing? Wait, no.

Fourth option: a parabola opening downward, with vertex at (0, 4), crossing x-axis at x=-2 and x=2.

Wait, let's think about the original function's derivative:

Original function:

- For \( x < - 4 \): decreasing (slope negative) → derivative negative.
- Between \( - 4 \) and \( 2 \): Wait, no, at x=0, the original function has a slope (derivative) that is positive? Wait, at x=0, the original graph is going up (from x=0 to x=2, it's increasing), so slope at x=0 is positive. Wait, maybe the original function's derivative is a quadratic function (parabola) opening downward, because the original function has a single local maximum (at x=2) and maybe a local minimum? Wait, no, the original graph: left side (x < -4) is decreasing, then between -4 and 2, is it increasing? Wait, at x=-4, the original graph crosses the x-axis, then goes down to a minimum? Wait, maybe I need to re - examine.

Wait, the key is: the original function has a local maximum at x = 2 (where derivative is 0), and let's see the slope before and after:

- Before x = 2 (for x < 2), the original function is increasing (slope positive) → derivative positive.
- After x = 2 (for x > 2), the original function is decreasing (slope negative) → derivative negative.

Also, on the left side (x < -4), the original function is decreasing (slope negative) → derivative negative. Wait, but that can't be. Wait, maybe the original function is symmetric? No, the original graph: left side (x < -4) is going down, then between -4 and 2, it's going up (increasing), then between 2 and 4, going down (decreasing). So the derivative should be:

- For x < -4: slope negative (derivative negative)
- Between -4 and 2: slope positive (derivative positive)
- Between 2 and 4: slope negative (derivative negative)
- For x > 4: slope negative (derivative negative)? Wait, no, the original graph on the right (x > 3) goes down, so slope negative.

But the derivative graph should have:

- A zero at x = 2 (where original has a maximum, derivative zero).
- Also, maybe another zero? Wait, the original graph on the left: at x = -4, it crosses the x-axis, but is that a minimum? If the original function has a minimum at x = -4, then derivative would be zero there. So the derivative would have zeros at x = -4 and x = 2, and be a quadratic (parabola) opening downward (because the original function has a maximum at x=2 and a minimum at x=-4? Wait, no, if the derivative is a parabola opening downward, it would have a maximum between -4 and 2.

Wait, looking at the options:

The fourth option (the bottom one) is a parabola opening downward, with vertex at x=0, crossing the x-axis at x=-2 and x=2? No, wait, the fourth graph: x-intercepts at x=-2 and x=2, vertex at (0,4), opening downward.

Wait, maybe the original function's derivative is the fourth graph. Let's check:

- When x < -2: derivative negative (original function decreasing)
- Between -2 and 2: derivative positive (original function increasing)
- When x > 2: derivative negative (original function decreasing)

Which matches the original function's behavior:

- For x < -2: original function decreasing (slope negative)
- Between -2 and 2: original function increasing (slope positive)
- For x > 2: original function decreasing (slope negative)

Yes, that makes sense. So the derivative graph is the fourth one (the parabola opening downward, vertex at (0,4), crossing x-axis at x=-2 and x=2).

Step2: Match with the options

Looking at the options, the fourth graph (the one at the bottom, with peak at x=0, crossing x-axis at x=-2 and x=2, opening downward) is the derivative of the original function.

Answer:

The graph of the derivative is the fourth one (the parabola - shaped graph opening downward with vertex at \( x = 0 \), \( y = 4 \) and x - intercepts at \( x=-2 \) and \( x = 2 \)). (Assuming the fourth graph is the one with the peak at x=0, crossing x - axis at x=-2 and x=2, opening downward)